A260683 Number of 2's in the expansion of 2^n in base 3.

0, 1, 0, 2, 1, 1, 1, 2, 0, 4, 2, 4, 3, 3, 2, 6, 5, 5, 3, 7, 4, 7, 5, 4, 1, 5, 2, 8, 8, 7, 9, 9, 8, 7, 7, 8, 4, 6, 8, 9, 11, 11, 7, 11, 10, 8, 9, 8, 8, 10, 11, 16, 13, 10, 9, 12, 13, 16, 12, 13, 15, 15, 11, 15, 16, 14, 14, 12, 14, 15, 14, 16, 11, 18, 11, 17, 10

Offset

0,4

Comments

Erdős conjectures that a(n) > 0 for n > 8.

References

R. K. Guy, Unsolved Problems in Number Theory, B33. [Does not seem to be in section B33.]

Links

Robert Israel, Table of n, a(n) for n = 0..10000

Paul Erdős, Some unconventional problems in number theory, Mathematics Magazine, Vol. 52, No. 2 (1979), pp. 67-70.

Formula

a(n) = A020915(n) - A104320(n) - A036461(n). - Altug Alkan, Nov 15 2015

a(n) = A081603(A000079(n)). - Michel Marcus, Dec 03 2015

Example

For n=5, the expansion of 2^n in number base 3 is 1012, thus: a(n)=1
For n=10, the expansion of 2^n in number base 3 is 1101221, thus: a(n)=2

Maple

seq(numboccur(2, convert(2^n, base, 3)), n=0..100); # Robert Israel, Nov 15 2015

Mathematica

S={}; n=-1; While[n<150, n++; A=IntegerDigits[2^n, 3]; k=Count[A, 2]; AppendTo[S, k]]; S

Prog

(PARI) c(k, d, b) = {my(c=0, f); while (k>b-1, f=k-b*(k\b); if (f==d, c++); k\=b); if (k==d, c++); return(c)}
for(n=0, 300, print1(c(2^n, 2, 3)", ")) \\ Altug Alkan, Nov 15 2015
(PARI) a(n) = #select(x->(x==2), digits(2^n, 3)); \\ Michel Marcus, Nov 28 2018
(Perl) use ntheory ":all"; sub a260683 { scalar grep { $_==2 } todigits(vecprod((2) x shift), 3) } # Dana Jacobsen, Aug 16 2016

Crossrefs

Cf. A004642 (2^n in base 3), A020915 (number of terms), A036461 (number of 1's), A104320 (number of 0's).

Cf. A000108 (conjecture that A000108(n) is 6m+1 only for n = 0, 1 and 5 follows from Erdős's one).

Cf. A005836 (for numbers with no 2 in base 3).

Sequence in context: A163819 A301734 A281185 * A337683 A362451 A092673

Adjacent sequences: A260680 A260681 A260682 * A260684 A260685 A260686

Keyword

base,easy,nonn

Author

Emmanuel Vantieghem, Nov 15 2015

Status

approved